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js 取时间差去掉周六周日实现代码

作者: 来源:www.28hudong.com 2013-03-30 00:17:12 阅读 我要评论

复制代码 代码如下: function GetDayNum(no,type) { if (type == "35") { var num7 = document.getElementById("6").value; var value = document.getElementById("7").value; var valueR = num7; if (no == "7") {valueR = value;}//标识时间值 if (num7 != "" && value != "") {//两个时间段可以融合 BUG1 var numValue = 0; var numValue = 0; var day = 24 * 60 * 60 * 1000; var dateArr = value.split("-"); var checkDate = new Date(); checkDate.setFullYear(dateArr[0], dateArr[1] - 1, dateArr[2]); var checkTime = checkDate.getTime(); var dateArr2 = num7.split("-"); var checkDate2 = new Date(); checkDate2.setFullYear(dateArr2[0], dateArr2[1] - 1, dateArr2[2]); var checkTime2 = checkDate2.getTime(); var numValue = (checkTime - checkTime2) / day; //相差的总天数 var totleWeek = numValue / 7; //差别多少周 var yuDay = numValue % 7; //除了整个星期的天数 var lastDay = 0; var startWeek = num7.replace(/-/g, "/"); //把“-”改为“/”,其中“-”算时间不准确这个我测试了 //var date = new Date(parseInt(startWeek[0]), parseInt(startWeek[1]), parseInt(startWeek[2])); var date = Date.parse(startWeek); var dateNew = new Date(date); var weekDay = dateNew.getDay(); //星期几 var endWeekDay = 0; //多余的天数有几天是周六或者周日 if ((weekDay == 6 && yuDay >= 2) || (weekDay == 7 && yuDay >= 1) || (weekDay == 5 && yuDay >= 3) || (weekDay == 4 && yuDay >= 4) || (weekDay == 3 && yuDay >= 5) || (weekDay == 2 && yuDay >= 6) || (weekDay == 1 && yuDay >= 7)) { endWeekDay = 2;www.2cto.com } if ((weekDay == 6 && yuDay < 1) || (weekDay == 7 && yuDay < 5) || (weekDay == 5 && yuDay < 2) || (weekDay == 4 && yuDay < 3) || (weekDay == 3 && yuDay < 4) || (weekDay == 2 && yuDay < 5) || (weekDay == 1 && yuDay < 6)) { endWeekDay = 1; } // if (numValue == 0) { numValue = 1; } numValue = numValue - (totleWeek * 2) - endWeekDay; //最终时间 } } }

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